If a is not equal to -1 then (1-a^16/(((1+a)(1+a^2)(1+a^4)(1+a^8)))=

In questions with an expression in the denominator, a GMAT type of question will always ensure that it gives you some information which will tell you that the denominator cannot be ZERO. In this question, that information is a is not equal to -1. Because if a = -1, (1+a) = 0 and the denominator will become ZERO. Since division by ZERO is not tested on the GMAT, a≠-1 is a necessary constraint.

Whenever you have an expression like (1-\(a^{16}\)) or even (\(a^{16}\) – 1), think of using the a^2\(\)-\(b^2\) identity to solve the question. In short, if you have a binomial with 1 as one of the terms and the other term having an even power, you need to apply the \(a^2\)-\(b^2\) identity (of course, the binomial should be a subtraction of two terms.

1-\(a^{16}\) = \(1^2\) – \((a^8)^2\) = (1-\(a^8\)) (1+\(a^8\)). At this stage, the (1+\(a^8\)) can be cancelled out and we are left with (1-\(a^8\)).

(1-\(a^8\)) = \(1^2\)- \((a^4)^2\) = (1-\(a^4\))(1+\(a^4\)). At this stage, (1+\(a^4\)) can be cancelled out and we are left with (1-\(a^4\)).

(1-\(a^4\)) = \(1^2\) – \((a^2)^2\) = (1-\(a^2\))(1+\(a^2\)). At this stage, (1+\(a^2\)) can be cancelled out and we are left with (1-\(a^2\)).

Clearly, (1-\(a^2\)) can be broken up as (1-a)(1+a). (1+a) cancels out with the (1+a) in the denominator and we are left with (1-a).

The correct answer option is D.

In this question, there were two conspicuous clues according to me – the expression with even powers in the numerator and the expressions in the denominator with reducing powers which were factors of 16. When I saw these two in the same place, I figured out that I had to break down the numerator and a lot of the terms in the denominator would cancel out.

Hope that helps!